Integrand size = 33, antiderivative size = 187 \[ \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {4 a (9 A+8 B) \tan (c+d x)}{45 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (9 A+8 B) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a B \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}-\frac {8 (9 A+8 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {4 (9 A+8 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 a d} \]
4/105*(9*A+8*B)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+4/45*a*(9*A+8*B)*tan (d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/63*a*(9*A+8*B)*sec(d*x+c)^3*tan(d*x+c)/ d/(a+a*sec(d*x+c))^(1/2)+2/9*a*B*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c) )^(1/2)-8/315*(9*A+8*B)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 0.41 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.52 \[ \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 a \left (16 (9 A+8 B)+8 (9 A+8 B) \sec (c+d x)+6 (9 A+8 B) \sec ^2(c+d x)+5 (9 A+8 B) \sec ^3(c+d x)+35 B \sec ^4(c+d x)\right ) \tan (c+d x)}{315 d \sqrt {a (1+\sec (c+d x))}} \]
(2*a*(16*(9*A + 8*B) + 8*(9*A + 8*B)*Sec[c + d*x] + 6*(9*A + 8*B)*Sec[c + d*x]^2 + 5*(9*A + 8*B)*Sec[c + d*x]^3 + 35*B*Sec[c + d*x]^4)*Tan[c + d*x]) /(315*d*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.99 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4504, 3042, 4290, 3042, 4287, 27, 3042, 4489, 3042, 4279}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a} (A+B \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4504 |
\(\displaystyle \frac {1}{9} (9 A+8 B) \int \sec ^4(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a B \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{9} (9 A+8 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a B \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 4290 |
\(\displaystyle \frac {1}{9} (9 A+8 B) \left (\frac {6}{7} \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{9} (9 A+8 B) \left (\frac {6}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 4287 |
\(\displaystyle \frac {1}{9} (9 A+8 B) \left (\frac {6}{7} \left (\frac {2 \int \frac {1}{2} \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{9} (9 A+8 B) \left (\frac {6}{7} \left (\frac {\int \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{9} (9 A+8 B) \left (\frac {6}{7} \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 4489 |
\(\displaystyle \frac {1}{9} (9 A+8 B) \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{9} (9 A+8 B) \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 4279 |
\(\displaystyle \frac {1}{9} (9 A+8 B) \left (\frac {6}{7} \left (\frac {\frac {14 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
(2*a*B*Sec[c + d*x]^4*Tan[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]]) + ((9*A + 8*B)*((2*a*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + (6*((2*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a*d) + ((14*a^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a*Sqrt[a + a*Sec[c + d*x]]*Ta n[c + d*x])/(3*d))/(5*a)))/7))/9
3.2.18.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2 ))), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b*( m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*d*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/( f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*a*d*((n - 1)/(b*(2*n - 1))) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; Fre eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B , e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b *(m + 1), 0] && !LtQ[m, -2^(-1)]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*C ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)) Int[Sqrt[a + b*Csc[e + f* x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ [A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && !LtQ[n, 0]
Time = 4.40 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.70
method | result | size |
default | \(\frac {2 \left (144 A \cos \left (d x +c \right )^{4}+128 B \cos \left (d x +c \right )^{4}+72 A \cos \left (d x +c \right )^{3}+64 B \cos \left (d x +c \right )^{3}+54 A \cos \left (d x +c \right )^{2}+48 B \cos \left (d x +c \right )^{2}+45 A \cos \left (d x +c \right )+40 B \cos \left (d x +c \right )+35 B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}\) | \(130\) |
parts | \(\frac {2 A \left (16 \cos \left (d x +c \right )^{3}+8 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+5\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{35 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B \left (128 \cos \left (d x +c \right )^{4}+64 \cos \left (d x +c \right )^{3}+48 \cos \left (d x +c \right )^{2}+40 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}\) | \(156\) |
2/315/d*(144*A*cos(d*x+c)^4+128*B*cos(d*x+c)^4+72*A*cos(d*x+c)^3+64*B*cos( d*x+c)^3+54*A*cos(d*x+c)^2+48*B*cos(d*x+c)^2+45*A*cos(d*x+c)+40*B*cos(d*x+ c)+35*B)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^3
Time = 0.26 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.65 \[ \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (16 \, {\left (9 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{4} + 8 \, {\left (9 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (9 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (9 \, A + 8 \, B\right )} \cos \left (d x + c\right ) + 35 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]
2/315*(16*(9*A + 8*B)*cos(d*x + c)^4 + 8*(9*A + 8*B)*cos(d*x + c)^3 + 6*(9 *A + 8*B)*cos(d*x + c)^2 + 5*(9*A + 8*B)*cos(d*x + c) + 35*B)*sqrt((a*cos( d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^5 + d*cos(d*x + c )^4)
\[ \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]
\[ \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{4} \,d x } \]
16/315*(315*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*((A*d*cos(2*d*x + 2*c)^4 + A*d*sin(2*d*x + 2*c)^4 + 4*A*d*cos(2 *d*x + 2*c)^3 + 6*A*d*cos(2*d*x + 2*c)^2 + 4*A*d*cos(2*d*x + 2*c) + 2*(A*d *cos(2*d*x + 2*c)^2 + 2*A*d*cos(2*d*x + 2*c) + A*d)*sin(2*d*x + 2*c)^2 + A *d)*integrate((((cos(12*d*x + 12*c)*cos(2*d*x + 2*c) + 5*cos(10*d*x + 10*c )*cos(2*d*x + 2*c) + 10*cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 10*cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 5*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(12*d*x + 12*c)*sin(2*d*x + 2*c) + 5*sin(10*d*x + 10*c)*sin(2 *d*x + 2*c) + 10*sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 10*sin(6*d*x + 6*c)*s in(2*d*x + 2*c) + 5*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2 )*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c) *sin(12*d*x + 12*c) + 5*cos(2*d*x + 2*c)*sin(10*d*x + 10*c) + 10*cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 10*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 5*cos(2* d*x + 2*c)*sin(4*d*x + 4*c) - cos(12*d*x + 12*c)*sin(2*d*x + 2*c) - 5*cos( 10*d*x + 10*c)*sin(2*d*x + 2*c) - 10*cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 1 0*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 5*cos(4*d*x + 4*c)*sin(2*d*x + 2*c)) *sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin (2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*sin(12*d*x + 12 *c) + 5*cos(2*d*x + 2*c)*sin(10*d*x + 10*c) + 10*cos(2*d*x + 2*c)*sin(8*d* x + 8*c) + 10*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 5*cos(2*d*x + 2*c)*si...
\[ \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{4} \,d x } \]
Time = 22.63 (sec) , antiderivative size = 512, normalized size of antiderivative = 2.74 \[ \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,16{}\mathrm {i}}{5\,d}+\frac {\left (48\,A-32\,B\right )\,1{}\mathrm {i}}{105\,d}\right )+\frac {\left (336\,A+672\,B\right )\,1{}\mathrm {i}}{105\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,16{}\mathrm {i}}{7\,d}-\frac {B\,320{}\mathrm {i}}{63\,d}\right )+\frac {B\,32{}\mathrm {i}}{7\,d}+\frac {\left (144\,A+288\,B\right )\,1{}\mathrm {i}}{63\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (-\frac {A\,16{}\mathrm {i}}{9\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,16{}\mathrm {i}}{9\,d}-\frac {\left (16\,A+32\,B\right )\,1{}\mathrm {i}}{9\,d}\right )+\frac {\left (16\,A+32\,B\right )\,1{}\mathrm {i}}{9\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (288\,A+256\,B\right )\,1{}\mathrm {i}}{315\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )}-\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (144\,A+128\,B\right )\,1{}\mathrm {i}}{315\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )} \]
((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*16i)/(5*d) + ((48*A - 32*B)*1i)/(105*d)) + ((336*A + 672*B)*1 i)/(105*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x* 1i)*((A*16i)/(7*d) - (B*320i)/(63*d)) + (B*32i)/(7*d) + ((144*A + 288*B)*1 i)/(63*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1 i)*((A*16i)/(9*d) - ((16*A + 32*B)*1i)/(9*d)) - (A*16i)/(9*d) + ((16*A + 3 2*B)*1i)/(9*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) - ( exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)) ^(1/2)*(288*A + 256*B)*1i)/(315*d*(exp(c*1i + d*x*1i) + 1)) - (exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(144 *A + 128*B)*1i)/(315*d*(exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1))